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Little help for my parabola?

  1. May 20, 2009 #1
    See, there is an equation which is f(x) = x^2 - 5x + 6. On the points 2 and 3 of the x axis, they reach 0. So where is the vertex? I'm guessing its -.25, but just to make sure i had to ask. Well, our class seemed to have alot of it and our teacher, Ms. Knudsvig, didn't know.
     
    Last edited: May 20, 2009
  2. jcsd
  3. May 20, 2009 #2

    HallsofIvy

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    You mean that f(2)= 4- 10+ 6= 0 and f(3)= 9- 15+ 6= 0. Because this parabola has vertical axis, it vertex is exactly half way between 2 and 3, at 5/2. f(5/2)= 25/4- 25/2+ 6= 25/4- 50/4+ 24/4= -1/4= 0.25.

    You could also get that, without using the fact that the vertex is exactly halfway between to points of equal height, by completing the square:

    5/2= 5/2 and (5/2)^2= 25/4 so f(x)= x^2- 5x+ 25/4- 25/4+ 6= (x- 5/2)^2- 25/4+ 24/4= (x- 5/2)^2- 1/4.

    Since a square is never negative, that is always -1/4 plus something. When x= 5/2, (5/2- 5/2)^2- 1/4= -1/4. The lowest point, the vertex, is (5/2, -1/4).
     
  4. May 20, 2009 #3
    To find the vertex, all you have to do is write the equation in vertex form. Standard from for a parabola is ax^2+bx+c. Vertex form for a parabola is y= a(x-h)^2+k. Once you have the equation in this form, h and k are the coordinates for your vertex.
     
  5. May 22, 2009 #4
    How much do you (darkfire) know about calculus?
     
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